(B) Maximum acceleration occurs when the derivative (slope) of velocity
is greatest.
(B) The object changes direction only when velocity changes sign.
Velocity changes sign from negative to positive at t = 5.
(D) From the graph, f ′(2) = 3, and we are told the line passes through
(2,10). We therefore have f (x) 10 + 3(x − 2) = 3x + 4.
(C) At x = 1 and 3, f ′(x) = 0; therefore f has horizontal tangents.
For x < 1, f ′ > 0; therefore f is increasing.
For x > 1, f ′ < 0, so f is decreasing.
For x < 2, f ′ is decreasing, so f ′′ < 0 and the graph of f is concave
downward.
For x > 2, f ′ is increasing, so f ′′ > 0 and the graph of f is concave
upward.
(C) Note that .
(D) Only at S does the graph both rise and change concavity.
(E) Only at T is the tangent horizontal and the curve concave down.
